If $\operatorname{lcm}\left(x_1,x_2,\ldots,x_n\right) = k$, then there exist integers $q_1, q_2, \ldots, q_n$ such that $x_1q_1=x_2q_2=\cdots=x_nq_n = k$ and $\operatorname{gcd}(q_1,q_2,\ldots,q_n) = 1$.
edit(3/3/2019):
Let's say $LCM(x_1,...;x_n) = [x_1,...;x_n]\\GCD(x_1,...,x_n) = (x_1,...,x_n)$
I'm trying to prove
$[x_1,...;x_n;y_1;...;y_k]=[[x_1;...;x_n][y_1;...;y_k]]$ (n,k>1)
I've done the same thing with GCD using the below property of it:
$GCD(x_1,x_2,...x_n) = d$ then:
\begin{cases}
x_1=dq_1;...;x_n=dq_n
\\
(q_1;q_2;...;q_n)=1
\end{cases}
Here's my proof of $(x_1,...,x_n;...;y_1;...y_k)=((x_1,...,x_n);(y_1;...;y_k))$
Assume $(x_1,...,x_n;...;y_1;...y_k) = d$,we then have:
\begin{cases}
x_1=du_1;...;x_n=du_n
\\
y_1=dv_1;...;y_k=dv_k
\\
(u_1;...;v_k) = 1
\end{cases}
Hence:
\begin{align}((x_1,...,x_n);(y_1;...;y_k))&=(d.(u_1;...;u_n);d.(v_1;...;v_k))\\
&=d.((u_1;...;u_n);(v_1;...;v_k))
&=d.1 = d (proved)
\end{align}
(We can easily prove ((u_1;...;u_n);(v_1;...;v_n)) = 1)
I intend to use the same method to prove the LCM one but can't find any appropriate properties of LCM to do that?Any ideas on my problem?